In fig. (6), the spring has a force constant k The pulley
 

s light and smooth, the spring and string are light. The suspended block has a mass m. If the block is slightly displaced from its equilibrium position and then released, the period of its vertical oscillation is

detailed solution

Correct option is B

see fig (12)Let the tension in spring at equilibrium position for block beT0. Then mg=T0 From force diagram of pulley2T0=ky0  or  2mg=ky0 When the block is displaced through a small distanceyin downward direction from equilibrium position, let the tension in string be T. The stretching of string will be (y/2). Therefore  and 2T=ky0+y02=ky0+ky02 From eqs. (1) and 2), we get 2T=2mg+ky02  or  T=mg+ky04 Now mg−mg+ky04=ma or ma=−ky04  or  a=−k4my0 therofore d2y0dt2=−k4my0  or  d2y0dt2+k4my0=0 T=2πω=2π4mR